Find the sum of all real solutions to the equation \[\frac{x-2}{x^2+4x+1} = \frac{x-5}{x^2-10x}.\]
Answer: To eliminate the fractions, we multiply by $(x^2+4x+1)(x^2-10x)$ on both sides, giving \[(x-2)(x^2-10x) = (x-5)(x^2+4x+1).\]Expanding both sides yields \[x^3 - 12x^2 + 20x = x^3 -x^2 -19x -5,\]and so \[0 =11x^2 -39 x -5.\]By Vieta's formulas, the sum of the roots of this equation is $\boxed{\tfrac{39}{11}}\,.$ (One can compute the roots explicitly and check that they do not make any of the denominators of the original equation equal to zero.)